Loops Infinity of a spectrum
Clash Royale CLAN TAG#URR8PPP
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Let $mathbfX$ be an (orthogonal) spectrum (can assume that it's an $Omega$-spectrum if this helps give a positive answer) and give the category of orthogonal spectra the stable model structure.
Let $Omega^infty$ be the total right derived functor of evaluation at $0$.
I want to show that
$$
Omega^infty (S^n wedge mathbfX) simeq mathbfX(BbbR^n).
$$
Attempt:
Since we can assume that $mathbfX$ is an $Omega$-spectrum, (hence fibrant in the stable model structure) $Omega^infty$ is simply just evaluation at $0$. So the left-hand side becomes
$$
S^n wedge mathbfX(0).
$$
This comes with an obvious structure map to $mathbfX(BbbR^n)$, but this map, in general, is not a weak homotopy equivalence.
model-categories stable-homotopy-theory spectra
asked Sep 5 at 11:05
274072
1215
 |Â
show 2 more comments
up vote
0
down vote
favorite
Let $mathbfX$ be an (orthogonal) spectrum (can assume that it's an $Omega$-spectrum if this helps give a positive answer) and give the category of orthogonal spectra the stable model structure.
Let $Omega^infty$ be the total right derived functor of evaluation at $0$.
I want to show that
$$
Omega^infty (S^n wedge mathbfX) simeq mathbfX(BbbR^n).
$$
Attempt:
Since we can assume that $mathbfX$ is an $Omega$-spectrum, (hence fibrant in the stable model structure) $Omega^infty$ is simply just evaluation at $0$. So the left-hand side becomes
$$
S^n wedge mathbfX(0).
$$
This comes with an obvious structure map to $mathbfX(BbbR^n)$, but this map, in general, is not a weak homotopy equivalence.
model-categories stable-homotopy-theory spectra
asked Sep 5 at 11:05
274072
1215
1
I believe that $(S^nwedge X)_0=(mathfrakL(0,mathbbR^n)wedge X)_0=mathfrakL(0,mathbbR^n)wedge X_0=(mathfrakL(0,-)wedge X_0)_ncong X_n$, since the last evaluation map is a component of a natural isomorphism.
– Tyrone
Sep 10 at 15:12
Thanks. Can you clarify the isomorphism?
– 274072
Sep 10 at 15:50
It is Lemma 4.9 in Equivariant Orthogonal Spectra and S-Modules by Mandell and May.
– Tyrone
Sep 10 at 16:11
1
Ok, my original argument for that seems less obvious than I thought. How about this, $Omega^infty(S^nwedge X)cong Map_*(S^0,Omega^infty(S^nwedge X))cong Sp(S^0,S^nwedge X)cong Sp(S^nwedge S^-n,S^nwedge X)cong Sp(S^-n,Omega^n(S^nwedge X))simeq Sp(S^-n,X)simeq X_n$, where I have used Lemma 3.8 of that book.
– Tyrone
Sep 11 at 10:07
1
Yes, I'm identifying $S^n$ with $Sigma^n S^ncong mathbbSwedge S^n$. $S^-n$ is defined to be the spectrum $(mathbbR^n)^*$ (in Mandell-May's notation this is definition 4.6 and notation 4.7) so the last isomorphism is Yoneda.
– Tyrone
Sep 11 at 12:47
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $mathbfX$ be an (orthogonal) spectrum (can assume that it's an $Omega$-spectrum if this helps give a positive answer) and give the category of orthogonal spectra the stable model structure.
Let $Omega^infty$ be the total right derived functor of evaluation at $0$.
I want to show that
$$
Omega^infty (S^n wedge mathbfX) simeq mathbfX(BbbR^n).
$$
Attempt:
Since we can assume that $mathbfX$ is an $Omega$-spectrum, (hence fibrant in the stable model structure) $Omega^infty$ is simply just evaluation at $0$. So the left-hand side becomes
$$
S^n wedge mathbfX(0).
$$
This comes with an obvious structure map to $mathbfX(BbbR^n)$, but this map, in general, is not a weak homotopy equivalence.
model-categories stable-homotopy-theory spectra
asked Sep 5 at 11:05
274072
1215
Let $mathbfX$ be an (orthogonal) spectrum (can assume that it's an $Omega$-spectrum if this helps give a positive answer) and give the category of orthogonal spectra the stable model structure.
Let $Omega^infty$ be the total right derived functor of evaluation at $0$.
I want to show that
$$
Omega^infty (S^n wedge mathbfX) simeq mathbfX(BbbR^n).
$$
Attempt:
Since we can assume that $mathbfX$ is an $Omega$-spectrum, (hence fibrant in the stable model structure) $Omega^infty$ is simply just evaluation at $0$. So the left-hand side becomes
$$
S^n wedge mathbfX(0).
$$
This comes with an obvious structure map to $mathbfX(BbbR^n)$, but this map, in general, is not a weak homotopy equivalence.
model-categories stable-homotopy-theory spectra
model-categories stable-homotopy-theory spectra
asked Sep 5 at 11:05
274072
1215
asked Sep 5 at 11:05
274072
1215
asked Sep 5 at 11:05
274072
1215
asked Sep 5 at 11:05
274072
1215
asked Sep 5 at 11:05
274072
1215
1215
1
I believe that $(S^nwedge X)_0=(mathfrakL(0,mathbbR^n)wedge X)_0=mathfrakL(0,mathbbR^n)wedge X_0=(mathfrakL(0,-)wedge X_0)_ncong X_n$, since the last evaluation map is a component of a natural isomorphism.
– Tyrone
Sep 10 at 15:12
Thanks. Can you clarify the isomorphism?
– 274072
Sep 10 at 15:50
It is Lemma 4.9 in Equivariant Orthogonal Spectra and S-Modules by Mandell and May.
– Tyrone
Sep 10 at 16:11
1
Ok, my original argument for that seems less obvious than I thought. How about this, $Omega^infty(S^nwedge X)cong Map_*(S^0,Omega^infty(S^nwedge X))cong Sp(S^0,S^nwedge X)cong Sp(S^nwedge S^-n,S^nwedge X)cong Sp(S^-n,Omega^n(S^nwedge X))simeq Sp(S^-n,X)simeq X_n$, where I have used Lemma 3.8 of that book.
– Tyrone
Sep 11 at 10:07
1
Yes, I'm identifying $S^n$ with $Sigma^n S^ncong mathbbSwedge S^n$. $S^-n$ is defined to be the spectrum $(mathbbR^n)^*$ (in Mandell-May's notation this is definition 4.6 and notation 4.7) so the last isomorphism is Yoneda.
– Tyrone
Sep 11 at 12:47
 |Â
show 2 more comments
1
I believe that $(S^nwedge X)_0=(mathfrakL(0,mathbbR^n)wedge X)_0=mathfrakL(0,mathbbR^n)wedge X_0=(mathfrakL(0,-)wedge X_0)_ncong X_n$, since the last evaluation map is a component of a natural isomorphism.
– Tyrone
Sep 10 at 15:12
Thanks. Can you clarify the isomorphism?
– 274072
Sep 10 at 15:50
It is Lemma 4.9 in Equivariant Orthogonal Spectra and S-Modules by Mandell and May.
– Tyrone
Sep 10 at 16:11
1
Ok, my original argument for that seems less obvious than I thought. How about this, $Omega^infty(S^nwedge X)cong Map_*(S^0,Omega^infty(S^nwedge X))cong Sp(S^0,S^nwedge X)cong Sp(S^nwedge S^-n,S^nwedge X)cong Sp(S^-n,Omega^n(S^nwedge X))simeq Sp(S^-n,X)simeq X_n$, where I have used Lemma 3.8 of that book.
– Tyrone
Sep 11 at 10:07
1
Yes, I'm identifying $S^n$ with $Sigma^n S^ncong mathbbSwedge S^n$. $S^-n$ is defined to be the spectrum $(mathbbR^n)^*$ (in Mandell-May's notation this is definition 4.6 and notation 4.7) so the last isomorphism is Yoneda.
– Tyrone
Sep 11 at 12:47
1
1
I believe that $(S^nwedge X)_0=(mathfrakL(0,mathbbR^n)wedge X)_0=mathfrakL(0,mathbbR^n)wedge X_0=(mathfrakL(0,-)wedge X_0)_ncong X_n$, since the last evaluation map is a component of a natural isomorphism.
– Tyrone
Sep 10 at 15:12
I believe that $(S^nwedge X)_0=(mathfrakL(0,mathbbR^n)wedge X)_0=mathfrakL(0,mathbbR^n)wedge X_0=(mathfrakL(0,-)wedge X_0)_ncong X_n$, since the last evaluation map is a component of a natural isomorphism.
– Tyrone
Sep 10 at 15:12
Thanks. Can you clarify the isomorphism?
– 274072
Sep 10 at 15:50
Thanks. Can you clarify the isomorphism?
– 274072
Sep 10 at 15:50
It is Lemma 4.9 in Equivariant Orthogonal Spectra and S-Modules by Mandell and May.
– Tyrone
Sep 10 at 16:11
It is Lemma 4.9 in Equivariant Orthogonal Spectra and S-Modules by Mandell and May.
– Tyrone
Sep 10 at 16:11
1
1
Ok, my original argument for that seems less obvious than I thought. How about this, $Omega^infty(S^nwedge X)cong Map_*(S^0,Omega^infty(S^nwedge X))cong Sp(S^0,S^nwedge X)cong Sp(S^nwedge S^-n,S^nwedge X)cong Sp(S^-n,Omega^n(S^nwedge X))simeq Sp(S^-n,X)simeq X_n$, where I have used Lemma 3.8 of that book.
– Tyrone
Sep 11 at 10:07
Ok, my original argument for that seems less obvious than I thought. How about this, $Omega^infty(S^nwedge X)cong Map_*(S^0,Omega^infty(S^nwedge X))cong Sp(S^0,S^nwedge X)cong Sp(S^nwedge S^-n,S^nwedge X)cong Sp(S^-n,Omega^n(S^nwedge X))simeq Sp(S^-n,X)simeq X_n$, where I have used Lemma 3.8 of that book.
– Tyrone
Sep 11 at 10:07
1
1
Yes, I'm identifying $S^n$ with $Sigma^n S^ncong mathbbSwedge S^n$. $S^-n$ is defined to be the spectrum $(mathbbR^n)^*$ (in Mandell-May's notation this is definition 4.6 and notation 4.7) so the last isomorphism is Yoneda.
– Tyrone
Sep 11 at 12:47
Yes, I'm identifying $S^n$ with $Sigma^n S^ncong mathbbSwedge S^n$. $S^-n$ is defined to be the spectrum $(mathbbR^n)^*$ (in Mandell-May's notation this is definition 4.6 and notation 4.7) so the last isomorphism is Yoneda.
– Tyrone
Sep 11 at 12:47
 |Â
show 2 more comments
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1
I believe that $(S^nwedge X)_0=(mathfrakL(0,mathbbR^n)wedge X)_0=mathfrakL(0,mathbbR^n)wedge X_0=(mathfrakL(0,-)wedge X_0)_ncong X_n$, since the last evaluation map is a component of a natural isomorphism.
– Tyrone
Sep 10 at 15:12
Thanks. Can you clarify the isomorphism?
– 274072
Sep 10 at 15:50
It is Lemma 4.9 in Equivariant Orthogonal Spectra and S-Modules by Mandell and May.
– Tyrone
Sep 10 at 16:11
1
Ok, my original argument for that seems less obvious than I thought. How about this, $Omega^infty(S^nwedge X)cong Map_*(S^0,Omega^infty(S^nwedge X))cong Sp(S^0,S^nwedge X)cong Sp(S^nwedge S^-n,S^nwedge X)cong Sp(S^-n,Omega^n(S^nwedge X))simeq Sp(S^-n,X)simeq X_n$, where I have used Lemma 3.8 of that book.
– Tyrone
Sep 11 at 10:07
1
Yes, I'm identifying $S^n$ with $Sigma^n S^ncong mathbbSwedge S^n$. $S^-n$ is defined to be the spectrum $(mathbbR^n)^*$ (in Mandell-May's notation this is definition 4.6 and notation 4.7) so the last isomorphism is Yoneda.
– Tyrone
Sep 11 at 12:47